I have the below example:
SELECT name, age, location, SUM(pay)
FROM employee
GROUP BY location
This as expected will give me an error:
ORA-00979: not a GROUP BY expression
How can I get around this? I need to group by one maybe two columns but need to return all columns even if they're not used in the GROUP BY clause, I've looked at sub-queries to get around it but have had no luck so far.
You can use analytic functions:
SELECT name
, age
, location
, pay
, SUM(pay) over (partition by location order by location ) total
FROM employee
So, you can return all rows even if they are not used in the grouping.
So you want to know the total pay by location, and you want to know the names and ages of employees at each location? How about:
SELECT e.NAME,
e.AGE,
e.LOCATION,
t.TOTAL_LOCATION_PAY
FROM EMPLOYEE e
INNER JOIN (SELECT LOCATION,
SUM(PAY) AS TOTAL_LOCATION_PAY
FROM EMPLOYEE
GROUP BY LOCATION) t
ON (t.LOCATION = e.LOCATION)
Share and enjoy.
(Group b[http://docs.oracle.com/javadb/10.6.2.1/ref/rrefsqlj32654.html] Must have an aggregate function in every column that is not in the group by clause. When you are grouping, means that you want one row per group. Distinct values of the columns in the clause appear in the final result set.
This is because oracle can't know which of the values for the column that you don't have in the group by to retrieve. Consider this:
A X
B X
Select col1, col2 from myTable group by col2; -- incorrect
Select min(col1), col2 from myTable group by col2; -- correct
Why is the first incorrect? Because oracle can't know whether to retrieve A or B for the X value you have to specify it. i.e. MIN, MAX, etc.
There is an alternative to this named analytic functions that allow you to work under windows of your result set.
Now if you want total employee pay by location, and every employee you may want this.
SELECT name, age, location, SUM(pay) OVER(PARTITION BY location)
FROM employee
I believe this is better than #Bob Jarvis query as you only query the table once. Please correct me if I'm wrong. He also has employees and employee. Typo?
Related
How do I create a COUNT column to count the repetitive values?
And I want to keep the table EXACTLY as below but add the last column (count_id).
The values at the left come from a JOIN so they are "equal".
Thanks! (I tried a lot)
You just want count(*) as a window function:
select t.*,
count(*) over (partition by id, name, department) as count_id
from t;
I'm using DB2 for a project and looking to find which Group has the fewest members without using the min feature. My idea is to find all the groups and then subtract out any group which has more members from some other group thus leaving me with the group with that has no more members than any other group, i.e. the min.
So far I have
SELECT DISTINCT P.group as Group, count(P.id) as Count
FROM People P
EXCEPT
SELECT P.group, count(P.id)
FROM People P, People O
WHERE count(P.cid) > count(O.cid);
With a schema for People like
create table People (
group varchar(25) not null,
id smallint not null,
);
I am getting the following error:
SQL0119N An expression starting with "CLUB" specified in a SELECT clause,
HAVING clause, or ORDER BY clause is not specified in the GROUP BY clause or
it is in a SELECT clause, HAVING clause, or ORDER BY clause with a column
function and no GROUP BY clause is specified. SQLSTATE=42803
If you could help point out what I am doing wrong or the correct format for such a query it would be greatly appreciated!
find which Group has the fewest members
You can aggregate by group, order by member count, and fetch the top row only:
select p.group as grp, count(*) as cnt
from people p
group by p.group
order by count(*)
fetch first 1 rows only
You should try to use min(). Very straight forward. From your Error message, it seems like your HAVING clause is wrong so it would look into that.
I have read in various websites about the count() function but I still cannot make this work.
I made a small table with (id, name, last name, age) and I need to retrieve all columns plus a new one. In this new column I want to display how many times a name shows up or repeats itself in the table.
I have made test and can retrieve but only COLUMN NAME with the count column, but I haven't been able to retrieve all data from the table.
Currently I have this
select a.n_showsup, p.*
from [test1].[dbo].[person] p,
(select count(*) n_showsup
from [test1].[dbo].[person])a
This gives me all data on output but on the column n_showsup it gives me just the number of rows, now I know this is because I'm missing a GROUP BY but then when I write group by NAME it shows me a lot of records. This is an example of what I need:
You can use window functions, if you RDBMS supports them:
select t.*, count(*) over(partition by name) n_showsup
from mytable t
Alternatively, you can join the table with an aggregation query that counts the number of occurences of each name:
select t.*, x.n_showsup
from mytable t
inner join (select name, count(*) n_showsup from mytable group by name) x
on x.name = t.name
While the window function approach (#GMB's answer) is the right way to go, thinking through this from a subquery approach (like you were headed towards) would look something like:
select p.*, a.n_showsup
from [test1].[dbo].[person] p
INNER JOIN (
select name, count(*) n_showsup
from [test1].[dbo].[person]
GROUP BY name
) a ON p.name = a.name
This is VERY close to what you had, the difference is that we are grouping that subquery by name (so we get a count by name) and we can use that in the join criteria which we do with the ON clause on that INNER JOIN.
You should really never ever use a comma in your FROM clause. Instead use a JOIN.
I realize this is an odd question, but I'd like to know if this is possible.
Let's say I have a DB with ages and IDs. I need to compare each ID's age to the average age, but I can't figure out how to do that without grouping or subqueries.
SELECT
ID,
AGE - AVG(AGE)
FROM
TABLE
I'm trying to get something like the above, but obviously that doesn't work because ID isn't grouped, but I group, then it calculates the average for each group, and not the table as a whole. How can I get a global average without a subquery?
SELECT ID,
AGE -
AVG(AGE) OVER (partition by ID) as age_2
FROM Table
I just read is global avg
SELECT ID,
AGE -
AVG(AGE) OVER () as age_2
FROM Table
The window logic for average age is:
SELECT ID, AGE - ( AVG(AGE) OVER () )
FROM TABLE;
You do not want ORDER BY in the partitioning clause.
How can I get a global average without a subquery?
Use a variable.
DECLARE #AvgAge decimal(4,3)
SELECT #AvgAge = AVG(AGE) FROM TABLE
SELECT ID, AGE - #AvgAge [AgeDiff] FROM TABLE
i would like your help with my query.I have a table employee.details with the following columns:
branch_name, firstname,lastname, age_float.
I want this query to list all the distinct values of the age_float
attribute, one in each row of the result table, and beside each in the second field show the
number of people in the details table who had ages less than or equal to that value.
Any ideas? Thank you!
You can use OLAP functions:
SELECT DISTINCT age_float,
COUNT(lastname) OVER(ORDER BY age_float) AS number
FROM employee_details
COUNT(lastname) OVER(ORDER BY age_float) AS number orders rows by age, and returns employees count whose age <= current row age
or a simple join:
SELECT A.age_float, count(lastname)
FROM (SELECT DISTINCT age_float FROM employee_details) A
JOIN employee_details AS ED ON ED.age_float <= A.age_float
GROUP BY A.age_float